12: Solubility Equilibria
Make sure you thoroughly understand the following essential ideas:
- Discuss the roles of lattice- and hydration energy in determining the solubility of a salt in water.
- Explain what a qualitative analysis separation scheme is, and how it works.
- Write the solubility product expression for a salt, given its formula.
- Explain the distinction between an ion product and a solubility product.
- Given the formula of a salt and its K s value, calculate the molar solubility .
- Explain the Le Chatelier principle leads to the common ion effect .
- Explain why a strong acid such as HCl will dissolve a sparingly soluble salt of a weak acid, but not a salt of a strong acid.
- Describe what happens (and why) when aqueous ammonia is slowly added to a solution of silver nitrate
Dissolution of a salt in water is a chemical process that is governed by the same laws of chemical equilibrium that apply to any other reaction. There are, however, a number of special aspects of of these equilibria that set them somewhat apart from the more general ones that are covered in the lesson set devoted specifically to chemical equilibrium. These include such topics as the common ion effect, the influence of pH on solubility, supersaturation, and some special characteristics of particularly important solubility systems.
Solubility: the dissolution of salts in water
Drop some ordinary table salt into a glass of water, and watch it "disappear". We refer to this as dissolution , and we explain it as a process in which the sodium and chlorine units break away from the crystal surface, get surrounded by H 2 O molecules, and become hydrated ions .
\[NaCl_{(s)} \rightarrow Na^+_{(aq)}+ Cl^–_{(aq)} \]
The designation (aq) means "aqueous" and comes from aqua , the Latin word for water. It is used whenever we want to emphasize that the ions are hydrated — that H 2 O molecules are attached to them.
Remember that solubility equilibrium and the calculations that relate to it are only meaningful when both sides (solids and dissolved ions) are simultaneously present. But if you keep adding salt, there will come a point at which it no longer seems to dissolve. If this condition persists, we say that the salt has reached its solubility limit , and the solution is saturated in NaCl. The situation is now described by
\[NaCl_{(s)} \rightleftharpoons Na^+_{(aq)}+ Cl^–_{(aq)}\]
in which the solid and its ions are in equilibrium .
Salt solutions that have reached or exceeded their solubility limits (usually 36-39 g per 100 mL of water) are responsible for prominent features of the earth's geochemistry. They typically form when NaCl leaches from soils into waters that flow into salt lakes in arid regions that have no natural outlets; subsequent evaporation of these brines force the above equilibrium to the left, forming natural salt deposits. These are often admixed with other salts, but in some cases are almost pure NaCl. Many parts of the world contain buried deposits of NaCl (known as halite) that formed from the evaporation of ancient seas, and which are now mined.
Solubilities are most fundamentally expressed in molar (mol L –1 of solution) or molal (mol kg –1 of water) units. But for practical use in preparing stock solutions, chemistry handbooks usually express solubilities in terms of grams-per-100 ml of water at a given temperature, frequently noting the latter in a superscript. Thus 6.9 20 means 6.9 g of solute will dissolve in 100 mL of water at 20° C. When quantitative data are lacking, the designations "soluble", "insoluble", "slightly soluble", and "highly soluble" are used. There is no agreed-on standard for these classifications, but a useful guideline might be that shown below.
The solubilities of salts in water span a remarkably large range of values, from almost completely insoluble to highly soluble. Moreover, there is no simple way of predicting these values, or even of explaining the trends that are observed for the solubilities of different anions within a given group of the periodic table.
Ultimately, the driving force for dissolution (and for all chemical processes) is determined by the Gibbs free energy change. But because many courses cover solubility before introducing free energy, we will not pursue this here. Dissolution of a salt is conceptually understood as a sequence of the two processes depicted above:
- breakup of the ionic lattice of the solid,
- followed by attachment of water molecules to the released ions.
The first step consumes a large quantity of energy, something that by itself would strongly discourage solubility. But the second step releases a large amount of energy and thus has the opposite effect. Thus the net energy change depends on the sum of two large energy terms (often approaching 1000 kJ/mol) having opposite signs. Each of these terms will to some extent be influenced by the size, charge, and polarizability of the particular ions involved, and on the lattice structure of the solid. This large number of variables makes it impossible to predict the solubility of a given salt. Nevertheless, there are some clear trends for how the solubilities of a series of salts of a given anion (such as hydroxides, sulfates, etc.) change with a periodic table group. And of course, there are a number of general solubility rules — for example, that all nitrates are soluble, while most sulfides are insoluble.
Solubility and temperature
Solubility usually increases with temperature - but not always. This is very apparent from the solubility-vs.-temperature plots shown here. (Some of the plots are colored differently in order to make it easier to distinguish them where they crowd together.) The temperature dependence of any process depends on its entropy change — that is, on the degree to which thermal kinetic energy can spread throughout the system. When a solid dissolves, its component molecules or ions diffuse into the much greater volume of the solution, carrying their thermal energy along with them. So we would normally expect the entropy to increase — something that makes any process take place to a greater extent at a higher temperature.
So why does the solubility of cerium sulfate (green plot) diminish with temperature? Dispersal of the Ce 3 + and SO 4 2– ions themselves is still associated with an entropy increase, but in this case the entropy of the water decreases even more owing to the ordering of the H 2 O molecules that attach to the Ce 3 + ions as they become hydrated. It's difficult to predict these effects, or explain why they occur in individual cases — but they do happen.
The Importance of Sparingly Soluble Solids
All solids that dissociate into ions exhibit some limit to their solubilities, but those whose saturated solutions exceed about 0.01 mol L –1 cannot be treated by simple equilibrium constants owing to ion-pair formation that greatly complicates their behavior. For this reason, most of what follows in this lesson is limited to salts that fall into the "sparingly soluble" category. The importance of sparingly soluble solids arises from the fact that formation of such a product can effectively remove the corresponding ions from the solution, thus driving the reaction to the right. Consider, for example, what happens when we mix solutions of strontium nitrate and potassium chloride in a 1:2 mole ratio. Although we might represent this process by
\[Sr(NO_3)_{2(aq)}+ 2 KCl_{(aq)}→ SrCl_{(aq)}+ 2 KNO_{3(aq)} \label{1}\]
the net ionic equation
\[Sr^{2+} + 2 NO_3^– + 2 K^+ + 2 Cl^– → Sr^{2+} + 2 NO_3^– + 2 K^+ + 2 Cl^–\]
indicates that no net change at all has taken place! Of course if the solution were than evaporated to dryness, we would end up with a mixture of the four salts shown in Equation \(\ref{1}\), so in this case we might say that the reaction is half-complete. Contrast this with what happens if we combine equimolar solutions of barium chloride and sodium sulfate:
\[BaCl_{2(aq)}+ Na_2SO_{4(aq)}→ 2 NaCl_{(aq)}+ BaSO_{4(s)} \label{2}\]
whose net ionic equation is
\[Ba^{2+} + \cancel{ 2 Cl^–} + \cancel{2 Na^+} + SO_4^{2–} → \cancel{2 Na^+} + \cancel{2 Cl^–} + BaSO_{4(s)}\]
which after canceling out like terms on both sides, becomes simply
\[Ba^{2+} + SO_4^{2– }→ BaSO_{4(s)} \label{3}\]
Because the formation of sparingly soluble solids is "complete" (that is, equilibria such as the one shown above for barium sulfate lie so far to the right), virtually all of one or both of the contributing ions are essentially removed from the solution. Such reactions are said to be quantitative , and they are especially important in analytical chemistry:
- Qualitative analysis: This most commonly refers to a procedural scheme, widely encountered in first-year laboratory courses, in which a mixture of cations (usually in the form of their dissolved nitrate salts) is systematically separated and identified on the basis of the solubilities of their various anion salts such as chlorides, carbonates, sulfates, and sulfides. Although this form of qualitative analysis is no longer employed by modern-day chemists (instrumental techniques such as atomic absorption spectroscopy are much faster and comprehensive), it is still valued as an educational tool for familiarizing students with some of the major classes of inorganic salts, and for developing basic skills relating to observing, organizing, and interpreting results in the laboratory.
- Quantitative gravimetric analysis: In this classical form of chemical analysis, an insoluble salt of a cation is prepared by precipitating it by addition of a suitable anion. The precipitate is then collected, dried, and weighed ("gravimetry") in order to determine the concentration of the cation in the sample. For example, a gravimetric procedure for determining the quantity of barium in a sample might involve precipitating the metal as the sulfate according to Equation \(\ref{3}\) above, using an excess of sulfate ion to ensure complete removal of the barium. This method of quantitative analysis became extremely important in the latter half of the nineteenth century, by which time reasonably accurate atomic weights had become available, and sensitive analytical balances had been developed. It was not until the 1960's that it became largely supplanted by instrumental techniques which were much quicker and accurate. Gravimetric analysis is still usually included as a part of more advanced laboratory instruction, largely as a means of developing careful laboratory technique.
Solubility Products and Equilibria
Some salts and similar compounds (such as some metal hydroxides) dissociate completely when they dissolve, but the extent to which they dissolve is so limited that the resulting solutions exhibit only very weak conductivities. In these salts, which otherwise act as strong electrolytes, we can treat the dissolution-dissociation process as a true equilibrium. Although this seems almost trivial now, this discovery, made in 1900 by Walther Nernst who applied the Law of Mass Action to the dissociation scheme of Arrhenius, is considered one of the major steps in the development of our understanding of ionic solutions.
Using silver chromate as an example, we express its dissolution in water as
\[Ag_2CrO_{4(s)} \rightarrow 2 Ag^+_{(aq)}+ CrO^{2–}_{4(aq)} \label{4a}\]
When this process reaches equilibrium ( which requires that some solid be present ), we can write (leaving out the " (aq) s" for simplicity)
\[Ag_2CrO_{4(s)} \rightleftharpoons 2 Ag^+ + CrO^{2–}_{4} \label{4b}\]
The equilibrium constant is formally
\[K = \dfrac{[Ag^+]^2[CrO_4^{2–}]}{[Ag_2CrO_{4(s)}]} = [Ag^+]^2[CrO_4^{2–}] \label{5a}\]
But because solid substances do not normally appear in equilibrium expressions, the equilibrium constant for this process is
\[[Ag^+]^2 [CrO_4^{2–}] = K_s = 2.76 \times 10^{–12} \label{5b}\]
Because equilibrium constants of this kind are written as products, the resulting K 's are commonly known as solubility products , denoted by \(K_s\) or \(K_{sp}\).
Strictly speaking, concentration units do not appear in equilibrium constant expressions. However, many instructors prefer that students show them anyway, especially when using solubility products to calculate concentrations. If this is done, \(K_s\) in Equation \(\ref{5b}\) would have units of mol 3 L –3 .
Equilibrium and non-equilibrium in solubility systems
An expression such as [Ag + ] 2 [CrO 4 2– ] in known generally as an ion product — this one being the ion product for silver chromate. An ion product can in principle have any positive value, depending on the concentrations of the ions involved. Only in the special case when its value is identical with K s does it become the solubility product. A solution in which this is the case is said to be saturated . Thus when
\[[Ag^+]^2 [CrO_4^{2–}] = 2.76 \times 10^{-12}\]
at the temperature and pressure at which this value \(K_s\) of applies, we say that the "solution is saturated in silver chromate".
A solution must be saturated to be in equilibrium with the solid. This is a necessary condition for solubility equilibrium, but it is not by itself sufficient . True chemical equilibrium can only occur when all components are simultaneously present. A solubility system can be in equilibrium only when some of the solid is in contact with a saturated solution of its ions. Failure to appreciate this is a very common cause of errors in solving solubility problems.
Undersaturated and supersaturated solutions
If the ion product is smaller than the solubility product, the system is not in equilibrium and no solid can be present. Such a solution is said to be undersaturated . A supersaturated solution is one in which the ion product exceeds the solubility product. A supersaturated solution is not at equilibrium, and no solid can ordinarily be present in such a solution. If some of the solid is added, the excess ions precipitate out and until solubility equilibrium is achieved.
How to know the saturation status of a solution
This is just a simple matter of comparing the ion product \(Q_s\) with the solubility product \(K_s\). S o for the system
\[Ag_2CrO_{4(s)} \rightleftharpoons 2 Ag^+ + CrO_4^{2–} \label{4ba}\]
a solution in which \(Q_s < K_s\) (i.e., \(K_s /Q_s > 1\)) is undersaturated (blue shading) and the no solid will be present. The combinations of [Ag + ] and [CrO 4 2– ] that correspond to a saturated solution (and thus to equilibrium) are limited to those described by the curved line. The pink area to the right of this curve represents a supersaturated solution.
A sample of groundwater that has percolated through a layer of gypsum (CaSO 4 , K s = 4.9E–5 = 10 –4.3 ) is found to have be 8.4E–5 M in Ca 2 + and 7.2E–5 M in SO 4 2– . What is the equilibrium state of this solution with respect to gypsum?
Solution
The ion product
\[Q_s = (8.4 \times 10^{–5})(7.2 \times 10^{-5}) = 6.0 \times 10^{–4}\]
exceeds \(K_s\), so the ratio K s / Q s > 1 and the solution is supersaturated in CaSO 4 .
How are solubilities determined?
There are two principal methods, neither of which is all that reliable for sparingly soluble salts:
- Evaporate a saturated solution of the solid to dryness, and weigh what's left.
- Measure the electrical conductivity of the saturated solution, which will be proportional to the concentrations of the ions.
How Solubilities relate to solubility products
The solubility (by which we usually mean the molar solubility ) of a solid is expressed as the concentration of the "dissolved solid" in a saturated solution. In the case of a simple 1:1 solid such as AgCl, this would just be the concentration of Ag + or Cl – in the saturated solution. However, for a more complicated stoichiometry such as as silver chromate, the solubility would be only one-half of the Ag + concentration.
For example, let us denote the solubility of Ag 2 CrO 4 as \(S\) mol L –1 . Then for a saturated solution, we have
- \([Ag^+] = 2S\)
- \( [CrO_4^{2–}] = S\)
Substituting this into Equation \(\ref{5b}\) above,
\[(2S)^2 (S) = 4S^3 = 2.76 \times 10^{–12}\]
\[S= \left( \dfrac{K_s}{4} \right)^{1/3} = (6.9 \times 10^{-13})^{1/3} = 0.88 \times 10^{-4} \label{6a}\]
thus the solubility is \(8.8 \times 10^{–5}\; M\).
Note that the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction. For this reason it is meaningless to compare the solubilities of two salts having the formulas \(A_2B\) and \(AB_2\), on the basis of their \(K_s\) values.
It is meaningless to compare the solubilities of two salts having different formulas on the basis of their \(K_s\) values.
under these conditions.
Solution
moles of solute in 100 mL; S = 0.0016 g / 78.1 g/mol = 2.05E-5 mol
S = 2.05E–5 mol/0.100 L = 2.05E-4 M
K s = [Ca 2 + ][F – ] 2 = ( S )(2 S ) 2 = 4 × (2.05E–4) 3 = 3.44E–11
Estimate the solubility of La(IO 3 ) 3 and calculate the concentration of iodate in equilibrium with solid lanthanum iodate, for which K s = 6.2 × 10 –12 .
Solution
The equation for the dissolution is
\[La(IO_3)_3 \rightleftharpoons La^{3+ }+ 3 IO_3^–\]
If the solubility is S , then the equilibrium concentrations of the ions will be
[La 3 + ] = S and [IO 3 – ] = 3 S . Then K s = [La 3+ ][IO 3 – ] 3 = S (3 S ) 3 = 27 S 4
27 S 4 = 6.2 × 10 –12 , S = ( ( 6.2 ÷ 27) × 10 –12 ) ¼ = 6.92 × 10 –4 M
[IO 3 – ] = 3 S = 2.08 × 10 –5 ( M )
Cadmium is a highly toxic environmental pollutant that enters wastewaters associated with zinc smelting (Cd and Zn commonly occur together in ZnS ores) and in some electroplating processes. One way of controlling cadmium in effluent streams is to add sodium hydroxide, which precipitates insoluble Cd(OH) 2 ( K s = 2.5E–14). If 1000 L of a certain wastewater contains Cd 2+ at a concentration of 1.6E–5 M , what concentration of Cd 2+ would remain after addition of 10 L of 4 M NaOH solution?
Solution
As with most real-world problems, this is best approached as a series of smaller problems, making simplifying approximations as appropriate.
Volume of treated water: 1000 L + 10 L = 1010 L
Concentration of OH – on addition to 1000 L of pure water:
(4 M ) × (10 L)/(1010 L) = .040 M
Initial concentration of Cd 2 + in 1010 L of water:
(1.6E–5 M ) x (100/101) ≈ 1.6E–5 M
The easiest way to tackle this is to start by assuming that a stoichiometric quantity of Cd(OH) 2 is formed — that is, all of the Cd 2 + gets precipitated.
| Concentrations | [Cd 2+ ], M | [OH – ], M |
|---|---|---|
| initial | 1.6E–5 | 0.04 |
| change | –1.6E–5 | –3.2E–5 |
| final: | 0 | 0.04 – 3.2E–5 ≈ .04 |
Now "turn on the equilibrium" — find the concentration of Cd 2 + that can exist in a 0.04 M OH – solution:
| Concentrations | [Cd 2+ ], M | [OH – ], M |
|---|---|---|
| initial | o | 0.04 |
| change | + x | +2 x |
| at equilibrium | x | 0.04 + 2 x ≈ .04 |
Substitute these values into the solubility product expression:
\[Cd(OH)_{2(s) } = [Cd^{2+}] [OH^–]^2 = 2.5 \times 10^{–14}\]
\[[Cd^{2+}] = \dfrac{2.5 \times 10^{–14}}{ 16 \times 10^{–4}} = 1.6 \times 10^{–13}\; M\]
Note that the effluent will now be very alkaline:
\[pH = 14 + \log 0.04 = 12.6\]
so in order to meet environmental standards an equivalent quantity of strong acid must be added to neutralize the water before it is released.
All Just a Simplification of Reality
The simple relations between K s and molar solubility outlined above, and the calculation examples given here, cannot be relied upon to give correct answers. Some of the reasons for this are explained in Part 2 of this lesson, and have mainly to do with incomplete dissociation of many salts and with complex formation in the presence of anions such as Cl – and OH – . The situation is nicely described in the article What Should We Teach Beginners about Solubility and Solubility Products? by Stephen Hawkes ( J Chem Educ. 1998 75(9) 1179-81). See also the earlier article by Meites, Pode and Thomas Are Solubilities and Solubility Products Related? ( J Chem Educ. 1966 43(12) 667-72).
It turns out that solubility equilibria more often than not involve many competing processes and their rigorous treatment can be quite complicated. Nevertheless, it is important that students master these over-simplified examples. However, it is also important that they are not taken too seriously!
The Common Ion Effect
It has long been known that the solubility of a sparingly soluble ionic substance is markedly decreased in a solution of another ionic compound when the two substances have an ion in common. This is just what would be expected on the basis of the Le Chatelier Principle; whenever the process
\[CaF_{2(s)} \rightleftharpoons Ca^{2+} + 2 F^– \label{7}\]
is in equilibrium, addition of more fluoride ion (in the form of highly soluble NaF) will shift the composition to the left, reducing the concentration of Ca 2 + , and thus effectively reducing the solubility of the solid. We can express this quantitatively by noting that the solubility product expression
\[[Ca^{2+}][F^–]^2 = 1.7 \times 10^{–10} \label{8}\]
must always hold, even if some of the ionic species involved come from sources other than CaF 2 (s) . For example, if some quantity x of fluoride ion is added to a solution initially in equilibrium with solid CaF 2 , we have
- \([Ca^{2+}] = S\)
- \([F^–] = 2S + x\)
so that
\[K_s = [Ca^{2+}][ F^–]^2 = S (2S + x)^2 . \label{9a}\]
\[K_s ≈ S x^2 \]
or
\[S ≈ \dfrac{K_s}{x^2} \label{9b}\]
University-level students should be able to derive these relations for ion-derived solids of any stoichiometry.
The plots shown below illustrate the common ion effect for silver chromate as the chromate ion concentration is increased by addition of a soluble chromate such as \(Na_2CrO_4\).
What's different about the plot on the right? If you look carefully at the scales, you will see that this one is plotted logarithmically (that is, in powers of 10.) Notice how a much wider a range of values can display on a logarithmic plot. The point of showing this pair of plots is to illustrate the great utility of log-concentration plots in equilibrium calculations in which simple approximations (such as that made in Equation \(\ref{9b}\) can yield straight-lines within the range of values for which the approximation is valid.
Calculate the solubility of strontium sulfate ( K s = 2.8 × 10 –7 ) in
- pure water and
- in a 0.10 mol L –1 solution of \(Na_2SO_4\).
\[S = \sqrt{K_s} = \sqrt{ 2.8 \times 10^{–7} } = 5.3 \times 10^{–4}\]
(b) In 0.10 mol L –1 Na 2 SO 4 , we have
= [Sr 2 + ][SO 4 2– ] = S × (0.10 + S ) = 2.8 × 10 –7Because S is negligible compared to 0.10 M, we make the approximation
= [Sr 2 + ][SO 4 2– ] ≈ S × (0.10 M) = 2.8 × 10 –7so
This is roughly 100 times smaller than the result from (a) .
Selective Precipitation and Separations
Differences in solubility are widely used to selectively remove one species from a solution containing several kinds of ions.
The solubility products of AgCl and Ag 2 CrO 4 are 1.8E–10 and 2.0E–12, respectively. Suppose that a dilute solution of AgNO 3 is added dropwise to a solution containing 0.001 M Cl – and 0.01 M CrO 4 2– .
- Which solid, AgCl or Ag 2 CrO 4 , will precipitate first?
- What fraction of the first anion will have been removed when the second just begins to precipitate? Neglect any volume changes.
Solution
The silver ion concentrations required to precipitate the two salts are found by substituting into the appropriate solubility product expressions:
- to precipitate AgCl: [Ag + ] = 1.8E-10 / .001 = 1.8E-7 M
- to precipitate Ag 2 CrO 4 : [Ag + ] = (2.0E-12 / .01) ½ = 1.4E–5 M
The first solid to form as the concentration of Ag + increases will be AgCl. Eventually the Ag + concentration reaches 1.4E-5 M and Ag 2 CrO 4 begins to precipitate. At this point the concentration of chloride ion in the solution will be 1.3E-5 M which is about 13% of the amount originally present.
The preceding example is the basis of the Mohr titration of chloride by Ag + , commonly done to determine the salinity of water samples. The equivalence point of this precipitation titration occurs when no more AgCl is formed, but there is no way of observing this directly in the presence of the white AgCl which is suspended in the container. Before the titration is begun, a small amount of K 2 CrO 4 is added to the solution. Ag 2 CrO 4 is red-orange in color, so its formation, which signals the approximate end of AgCl precipitation, can be detected visually.
Competing Equilibria involving solids
Solubility expression are probably the exception rather than the rule. Such equilibria are often in competition with other reactions with such species as H + or OH – , complexing agents, oxidation-reduction, formation of other sparingly soluble species or, in the case of carbonates and sulfites, of gaseous products. The exact treatments of these systems can be extremely complicated, involving the solution of large sets of simultaneous equations. For most practical purposes it is sufficient to recognize the general trends, and to carry out approximate calculations.
Salts of weak acids are soluble in strong acids, but strong acids will not dissolve salts of strong acids
The solubility of a sparingly soluble salt of a weak acid or base will depend on the pH of the solution. To understand the reason for this, consider a hypothetical salt MA which dissolves to form a cation M + and an anion A – which is also the conjugate base of a weak acid HA. The fact that the acid is weak means that hydrogen ions (always present in aqueous solutions) and M + cations will both be competing for the A – :
The weaker the acid HA, the more readily will reaction take place, thus gobbling up A – ions. If an excess of H + is made available by addition of a strong acid, even more A – ions will be consumed, eventually reversing reaction , causing the solid to dissolve.
In , for example, sulfate ions react with calcium ions to form insoluble CaSO 4 . Addition of a strong acid such as HCl (which is totally dissociated ) has no effect because CaCl 2 is soluble. Although H + can protonate some SO 4 2– ions to form hydrogen sulfate ("bisulfate") HSO 4 – , this ampholyte acid is too weak to reverse by drawing a significant fraction of sulfate ions out of CaSO 4 (s) .
Calculate the concentration of aluminum ion in a solution that is in equilibrium with aluminum hydroxide when the pH is held at 6.0.
The equilibria are
\[Al(OH)_3 \rightleftharpoons Al^{3+} + 3 OH^–\]
with
\[K_s = 1.4 \times 10^{–34}\]
and
\[H_2O \rightleftharpoons H^+ + OH^–\]
with
\[K_w = 1 \times 10^{–14}\]
Substituting the equilibrium expression for the second of these into that for the first, we obtain
\[[OH^–]^3 = \left( \dfrac{K_w}{ [H^+]}\right)^3 = \dfrac{K_s}{[Al^{3+}]}\]
(1.0 × 10 –14 ) / (1.0 × 10 –6 ) 3 = (1.4 × 10 –24 ) / [Al 3 + ]
from which we find
\[[Al^{3+}] = 1.4 \times 10^{–10}\; M\]
Formation of a Competing Precipitate
If two different anions compete with a single cation to form two possible precipitates, the outcome depends not only on the solubilities of the two solids, but also on the concentrations of the relevant ions.
These kinds of competitions are especially important in groundwaters, which acquire solutes from various sources as they pass through sediment layers having different compositions. As the following example shows, competing equilibria of these kinds are very important for understanding geochemical processes involving the formation and transformation of mineral deposits.
Suppose that groundwater containing 0.001 M F – and 0.0018 M CO 3 2– percolates through a sediment containing calcite, CaCO 3 . Will the calcite be replaced by fluorite, CaF 2 ?
The two solubility equilibria are
\[\ce{CaCO3 <=> Ca^{2+} + CO3^{2–} \quad K_s = 10^{–8.1}\]
\[\ce{CaF2 <=> Ca^{2+} + 2 F^{–} \quad K_s = 10^{–10.4}\]
Solution:
The equilibrium between the two solids and the two anions is
\[CaCO_3 + 2 F^–\rightleftharpoons CaF_2 + CO_3^{2–}\]
This is just the sum of the dissolution reaction for CaCO 3 and the reverse of that for CaF 2 , so the equilibrium constant is
\[K = \dfrac{[CO_3^{2–}]}{ [F^–]^2} = \dfrac{10^{–8.1}}{ 10^{–10.4}} = 200\]
That is, the two solids can coexist only if the reaction quotient Q ≤ 200. Substituting the given ion concentrations we find that
\[Q = \dfrac{0.0018}{0.0012} = 1800\]
Since Q > K , we can conclude that the calcite will not change into fluorite.
Complex Ion Formation
Most transition metal ions possess empty d orbitals that are sufficiently low in energy to be able to accept electron pairs from electron donors from cations, resulting in the formation of a covalently-bound complex ion . Even neutral species that have a nonbonding electron pair can bind to ions in this way. Water is an active electron donor of this kind, so aqueous solutions of ions such as Fe 3 + (aq) and Cu 2 + (aq) exist as the octahedral complexes Fe(H 2 O) 6 3+ and Cu(H 6 O) 6 2+ , respectively.
Many of the remarks made above about the relation between K s and solubility also apply to calculations involving complex formation. See Stephen Hawkes' article Complexation Calculations are Worse Than Useless ("... to the point of absurdity...and should not be taught" in introductory courses.) ( J Chem Educ. 1999 76(8) 1099-1100 ). However, it is very important that you understand the principles outlined in this section.
H 2 O is only one possible electron donor; NH 3 , CN – and many other species (known collectively as ligands ) possess lone pairs that can occupy vacant d orbitals on a metallic ion. Many of these bind much more tightly to the metal than does H 2 O, which will undergo displacement and substitution by one or more of these ligands if they are present in sufficiently high concentration.
If a sparingly soluble solid is placed in contact with a solution containing a ligand that can bind to the metal ion much more strongly than H 2 O, then formation of a complex ion will be favored and the solubility of the solid will be greater. Perhaps the most commonly seen example of this occurs when ammonia is added to a solution of copper(II) nitrate, in which the Cu 2 + (aq) ion is itself the complex hexaaquo complex ion shown at the left:
Because ammonia is a weak base, the first thing we observe is formation of a cloudy precipitate of Cu(OH) 2 in the blue solution. As more ammonia is added , this precipitate dissolves, and the solution turns an intense deep blue, which is the color of hexamminecopper(II) and the various other related species such as Cu(H 2 O) 5 (NH 3 ) 2+ , Cu(H 2 O) 4 (NH 3 ) 2 2+ , etc.
In many cases, the complexing agent and the anion of the sparingly soluble salt are identical. This is particularly apt to happen with insoluble chlorides, and it means that addition of chloride to precipitate a metallic ion such as Ag + will produce a precipitate at first, but after excess Cl – has been added the precipitate will redissolve as complex ions are formed.
Some important solubility systems
In this section, we discuss solubility equilibria that relate to some very commonly-encountered anions of metallic salts. These are especially pertinent to the kinds of separations that most college students are required to carry out (and understand!) in their first-year laboratory courses.
Solubility of oxides and hydroxides
Metallic oxides and hydroxides both form solutions containing OH – ions. For example, the solubilities of the [sparingly soluble] oxide and hydroxide of magnesium are represented by
\[Mg(OH)_{2(s)} → Mg^{2+} + 2 OH^– \label{10}\]
\[MgO_{(S)} + H_2O → Mg^{2+} + 2 OH^– \label{11}\]
If you write out the solubility product expressions for these two reactions, you will see that they are identical in form and value.
Recall that pH = –log 10 [H + ], so that [H + ] = 10 –pH .
One might naïvely expect that the dissolution of an oxide such as MgO would yield as one of its products the oxide ion O 2+ . But the oxide ion is such a strong base that it grabs a proton from water, forming two hydroxide ions instead:
\[O^{2+} + H_2O → 2 OH^–\]
This is an example of the rule that the hydroxide ion is the strongest base that can exist in aqueous solution. 2" is an equilibrium mixture of hydrated CO 2 molecules and carbonic acid. To keep things as simple as possible, we will not distinguish between them in what follows, and just use the formula H 2 CO 3 to represent the two species collectively.
The other Group 2 metals, especially Mg, along with iron and several other transition elements are also found in carbonate sediments. When rain falls through the air, it absorbs atmospheric carbon dioxide, a small portion of which reacts with the water to form carbonic acid. Thus all pure water in contact with the air becomes acidic, eventually reaching a pH of 5.6.
As noted above, the equilibrium between bicarbonate and carbonate ions depends on the pH. Since the pH scale is logarithmic, it makes sense (and greatly simplifies the construction of the plot) to employ a log scale for the concentrations. The plot shown below corresponds to a total carbonate-system concentration of 10 –3 M , which is representative of many ground waters. For river and lake waters, 10 –5 M would be more typical; this would simply shift the curves downward without affecting their shapes.
Points 1 and 2 where adjacent curves overlap correspond to the two pK's. Recall that when the pH is the same as the pK, the concentrations of the two conjugate species are identical and half of the total system concentration. This places the crossover points at log 0.5 = –0.3 below the system concentration level.
A 10 –3 M solution of sodium bicarbonate would have a pH denoted by point 3, with [H 2 CO 3 ] and [CO 3 2– ] constituting only 1% (10 –5 M ) of the system. This corresponds to the equilibrium
\[2 HCO_3^– \rightleftharpoons H_2CO_3 + CO_3^{2–}\]
Carbonates act as bases and, as such, react with acids. Thus, the portion of the global water cycle that transports carbon from the air into natural waters constitutes a gigantic acid-base reaction that yields hydrogen carbonate ions, commonly referred to as bicarbonate. The natural waters that result have pH values between 6 and 10 and are essentially solutions of bicarbonates.
Limestone caves and sinkholes
When rainwater permeates into the soil, it can become even more acidic owing to the additional CO 2 produced by soil organisms. Also, the deeper the water penetrates, the greater its hydrostatic pressure and the more CO 2 it can hold, further increasing its acidity. If this water then works its way down through the fissures and cracks within a limestone layer, it will dissolve some of limestone, leaving void spaces which may eventually grow into limestone caves or form sinkholes that can swallow up cars or houses.
A well-known feature of limestone caves is the precipitated carbonate formations that decorate the ceilings and floors. These are known as stalactites and stalagmites , respectively. When water emerges from the ceiling of a cave that is open to the atmosphere, some of the excess CO 2 it contains is released as it equilibrates with the air. This raises its pH and thus reduces the solubility of of the carbonates, which precipitate as stalactites. Some of the water remains supersaturated and does not precipitate until it drips to the cave floor, where it builds up the stalagmite formations.
Hard Water
This term refers to waters that, through contact with rocks and sediments in lakes, streams, and especially in soils (groundwaters), have acquired metallic cations such as Ca 2 + , Mg 2 + , Fe 2 + , Fe 3 + , Zn 2 + Mn 2 + , etc. Owing to the ubiquity of carbonate sediments, the compensating negative charge is frequently supplied by the bicarbonate ion HCO 3 – , but other anions such as SO 4 2– , F – , Cl – , PO 4 3– and SiO 4 2– may also be significant.
Solid bicarbonates are formed only by Group 1 cations and all are readily soluble in water. But because HCO 3 – is amphiprotic, it can react with itself to yield carbonate:
\[2 HCO_3^– → H_2O + CO_3^[2–} + CO_{2(g)}\]
If bicarbonate-containing water is boiled, the CO 2 is driven off, and the equilibrium shifts to the right, causing any Ca 2 + or similar ions to form a cloudy precipitate. If this succeeds in removing the "hardness cations", the water has been "softened". Such water is said to possess carbonate hardness , sometimes known as " temporary hardness ". Waters in which anions other than HCO 3 – predominate cannot be softened by boiling, and thus possess non-carbonate hardness or " permanent hardness ".
Hard waters present several kinds of problems, both in domestic and industrial settings:
- Waters containing dissolved salts leave solid deposits when they evaporate. Residents of areas having hard water (about 85 percent of the U.S.) notice evaporative deposits on shower walls, in teakettles, and on newly-washed windows, glassware, and vehicles.
- Much more seriously from an economic standpoint, evaporation of water in boilers used for the production of industrial steam leaves coatings on the heat exchanger surfaces that impede the transfer of heat from the combustion chamber, reducing the thermal transfer efficiency. The resultant overheating of these surfaces can lead to their rupture, and in the case of high-pressure boilers, to disastrous explosions. In the case of calcium and magnesium carbonates, the process is exacerbated by the reduced solubility of these salts at high temperatures. Removal of boiler scales is difficult and expensive.
- Municipal water supplies in hard-water areas tend to be supersaturated in hardness ions. As this water flows through distribution pipes and the plumbing of buildings, these ions often tend to precipitate out on their interior surfaces. Eventually, this scale layer can become thick enough to restrict or even block the flow of water through the pipes. When scale deposits within appliances such as dishwashers and washing machines, it can severely degrade their performance.
- Cations of Group 2 and above react with soaps, which are sodium salts of fatty acids such as stearic acid, C 17 H 35 COOH. The sodium salts of such acids are soluble in water, which allows them to dissociate and act as surfactants:
\[C_{17}H_{35}COONa → C_{17}H_{35}COO^– Na^+\]
but the presence of polyvalent ions causes them to form precipitates
\[2 C_{17}H_{35}COO^– + Ca^{2+} → (C_{17}H_{35}COO^–)_2Ca_{(s)}\]
Calcium stearate is less dense than water, so it forms a scum that floats on top of the water surface; anyone who lives in a hard-water area is likely familiar with the unsightly "bathtub rings" it leaves around the high-water mark or the shower-wall stains.
Solubility Complications
All heterogeneous equilibria, on close examination, are beset with complications. But solubility equilibria are somewhat special in that there are more of them. Back in the days when the principal reason for teaching about solubility equilibria was to prepare chemists to separate ions in quantitative analysis procedures, these problems could be mostly ignored. But now that the chemistry of the environment has grown in importance — especially that relating to the ocean and natural waters — there is more reason for chemical scientists to at least know about the limitations of simple solubility products. This section will offer a quick survey of the most important of these complications, while leaving their detailed treatment to more advanced courses.
Tabulated K s values are notoriously unreliable
Many of the \(K_s\) v alues found in tables were determined prior to 1940 (some go back to the 1880s!) at a time before highly accurate methods became av ailable. Especially suspect are many of those for highly insoluble salts which are more difficult to measure. A table showing the variations in \(K_{sp}\) values for the same salts among ten textbooks was published by Clark and Bonikamp in J Chem Educ. 1998 75(9) 1183-85.A good An example that used a variety of modern techniques to measure the solubility of silver chromate was published by A.L. Jones et al in the Australian J. of Chemistry, 1971 24 2005-12.
Generations of chemistry students have amused themselves by comparing the disparate K s values to be found in various textbooks and table. In some cases, they differ by orders of magnitude. There are several reasons for this in addition to the ones described in detail further on.
- The most direct methods of measuring solubilities tend to not be very accurate for sparingly soluble salts. Two-significant figure precision is about the best one can hope in a single measurement.
- Many insoluble salts can exist in more than one crystalline form ( polymorphs ), and in some cases also as amorphous solids. Precipitation under different conditions (in the presence of different ions, at different temperatures, etc.) can yield different or mixed polymorphs.
- Other ions present in the solution can often get incorporated into the crystalline solid, usually replacing an ion of similar size ( substitutional solid solutions ). When this happens, it is no longer valid to write the equilibrium condition as a simple "product". This is very common in mineral deposits, and an important consideration in geochemistry,
Most salts are not Completely Dissociated in Water
The dissolution of cadmium iodide is water is commonly represented as
\[CdI_{2(s)} → Cd^{2+} + 2 I^–\]
Firstly, they combine to form neutral, largely-covalent molecular species:
\[Cd^{2+}_{(aq)} + 2 I^–_{(aq)} → CdI_{2(aq)}\]
This non-ionic form accounts for 78% of the Cd present in the solution! In addition, they form a molecular ion \(CdI^–_{(aq)}\) according to the following scheme:
| \(CdI_{2(s)} \rightleftharpoons Cd^{2+} + 2 I^–\) | \(K_1 = 10^{–3.9}\) |
| \(Cd^{2+} + I^– \rightleftharpoons CdI^+\) | \(K_2= 10^{+2.3}\) |
| \(CdI2_{(s)} \rightleftharpoons CdI^++ I^–\) | \(K = 10^{–1.6} = 0.023\) |
The data shown Tables \(\PageIndex{1}\) and \(\PageIndex{2}\) are taken from the article Salts are Mostly NOT Ionized by Stephen Hawkes: 1996 J Chem Educ. 73(5) 421-423. This fact was stated by Arrhenius in 1887, but has been largely ignored and is rarely mentioned in standard textbooks.
As a consequence, the concentration of "free" Cd 2 + (aq) in an aqueous cadmium iodide solution is only about 2% of the value you would calculate by taking K 1 as the solubility product. The principal component of such as solution is actually [covalently-bound] CdI 2 (aq) . It turns out that many salts, especially those of metals beyond Group 2 , are similarly only partially ionized in aqueous solution:
| salt | molarity | % cation | other species |
|---|---|---|---|
| KCl | 0.52 | 95 | KCl (aq) 5% |
| MgSO 4 | 0.04 | 58 | MgSO 4 (aq) 42% |
| CaCl 2 | 0.44 | 70 | CaCl + (aq) 30% |
| CuSO 4 | 0.045 | 56 | CuSO 4 (aq) 44% |
| CdI 2 | 0.50 | 2 | CdI 2 (aq) 76%, CdI – (aq) 22% |
| FeCl 3 | 0.1 | 10 | FeCl 2 + (aq) 42%, FeCl 2 (aq) 40%, FeOH 2 + (aq) 6%, Fe(OH) 2 + (aq) 2% |
If you are enrolled in an introductory course and do not plan on taking more advanced courses in chemistry or biochemistry, you can probably be safe in ignoring this, since your instructor and textbook likely do so. However, if you expect to do more advanced work or teach, you really should take note of these points, since few textbooks mention them.
Formation of Hydrous Complexes
Transition metal ions form a large variety of complexes with H 2 O and OH – , both of which have electron-pairs available to coordinate with the central ion. This gives rise to a large variety of soluble species that are in competition with an insoluble solid. Because of this, a single equilibrium constant (solubility product) cannot describe the behavior of a solid such as Fe(OH) 3 , which we summarize here as an example.
Aquo complexes : The electrostatic field of the positively-charged metal ion enhances the acidic nature of these H 2 O molecules, encouraging them to shed a proton and leaving OH – groups in their place.
\[Fe(H_2O)_6^{3+} → Fe(H_2O)_5(OH)^{2+}+H^+\]
This is just the first of a series of similar reactions, each one having a successively smaller equilibrium constant:
\[Fe(H_2O)_5(OH)^{2+}→ Fe(H_2O)_4(OH)_2^+→ Fe(H_2O)_3(OH)_3 → Fe(H_2O)_2(OH)_4^-\]
Hydroxo complexes: But there's more: when the hydroxide ion acts as a ligand, it gives rise to a series of hydroxo complexes, of which the insoluble Fe(OH) 3 can be considered a member:
| Fe 3 + + 3 OH – → Fe(OH) 3 (s) | 1/ K s = 10 38 |
| Fe 3 + + H 2 O → Fe(OH) 2 + + H + | K = 10 –2.2 |
| Fe 3 + + 2H 2 O → FeOH + + 2H + | K = 10 –6.7 |
| Fe 3 + + 4H 2 O → Fe(OH) 4 – + 4H + | K = 10 –23 |
| 2Fe 3 + + 2H 2 O → Fe 2 (OH) 2 4+ + 2H + | K = 10 –2.8 |
The equilibria listed above all involve H + and OH – ions, and are therefore pH dependent, as illustrated by the straight lines in the plot, whose slopes reflect the pH dependence of the corresponding ionic species. At any given pH, the equilibrium with solid Fe(OH) 3 is controlled by the ionic species having the highest concentration at any given pH. The corresponding lines in the plot therefore delineate the region (indicated by the orange shading) at which the solid can exist.
Ionic interactions: The "non-common ion effect"
A sparingly-soluble salt will be more soluble in a solution that contains non-participating ions. This is just the opposite of the common ion effect, and it might at first seem rather counter-intuitive: why would adding more ions of any kind make a salt more likely to dissolve?
Figure \(\PageIndex{2}\): Solubility of thallium iodate in solutions containing dissolved salts
A clue to the answer can be found in another fact: the higher the charge of the foreign ion, the more pronounced is the effect. This tells us that inter-ionic (and thus electrostatic) interactions must play a role. The details are rather complicated, but the general idea is that all ions in solution, besides possessing tightly-held waters of hydration, tend to attract oppositely-charged ions (" counter-ions ") around them. This "atmosphere" of counterions is always rather diffuse, but much less so (and more tightly bound) when one or both kinds of ions have greater charges. From a distance, these ion-counterion bodies appear to be almost electrically neutral, which keeps them from interacting with each other (as to form a precipitate).
The overall effect is to reduce the concentrations of the less-shielded ions that are available to combine to form a precipitate. We say that the thermodynamically-effective concentrations of these ions are less than their "analytical" concentrations. Chemists refer to these effective concentrations as ionic activities , and they denote them by curly brackets {Ag + } as opposed to square brackets [Ag + ] which refer to the nominal or analytical concentrations.
Although the concentrations of ions in equilibrium with a sparingly soluble solid are so low that they are essentially the same as the activities, the presence of other ions at concentrations of about 0.001M or greater can materially reduce the activities of the dissolution products, permitting the solubilities to be greater than what simple equilibrium calculations would predict.
Measured solubilities are averages and depend on size
The heterogeneous nature of dissolution reactions leads to a number of peculiar effects relating to the nature of equilibria involving surfaces. These arise from the fact that the tendency of a crystalline solid to dissolve will depend on the particular face or location from which dissolution occurs. Since all crystals present a variety of faces to the solution, a measured K s is really an average of values for these various faces.
And because many salts can exhibit different external shapes depending on the conditions under which they are formed, solubility products are similarly dependent on these conditions.
Very small crystals are more soluble than big ones
Molecules or ions that are situated on edges or corners are less strongly bound to the remainder of the solid than those on plane surfaces, and will consequently tend to dissolve more readily. Thus the leftmost face in the schematic lattice below will have more edge-bound molecular units than the other two, and this face (11) will be more soluble.
This means, among other things, that smaller crystals, in which the ratio of edges and corners is greater, will tend to have greater K s values than larger ones. As a consequence, smaller crystals will tend to disappear in favor of larger ones. Practical use is sometimes made of this when the precipitate initially formed in a chemical analysis or separation is too fine to be removed by filtration. The suspension is held at a high temperature for several hours, during with time the crystallites grow in size. This procedure is sometimes referred to as digestion.
Formation of supersaturated solutions
Contrary to what you may have been taught, precipitates do not form when the ion concentration product reaches the solubility product of a salt in a solution that is pure and initially unsaturated; to form a precipitate from a homogeneous solution, a certain degree of supersaturation is required. The extent of supersaturation required to initiate precipitation can be surprisingly great. Thus formation of barium sulfate BaSO 4 by combining the two kinds of ions does not occur until Q s exceeds K s by a factor of 160 or more. In part, this reflects the fact that precipitation proceeds by a series of reactions beginning with formation of an ion-pair which eventually becomes an ion cluster:
Ba 2 + + SO 4 2– → (BaSO 4 ) 0 → (BaSO 4 ) 2 0 → (BaSO 4 ) 3 0 → etc.
Owing to their overall neutrality, these aggregates are not stabilized by hydration, so they are more likely to break up than not. But a few may eventually survive until they are large enough (but still submicroscopic in size) to serve as precipitation nuclei.
Many substances other than salts form supersaturated solutions, and some salts form them more readily than others. Supersaturated solutions are easily made by dissolving the solid to near its solubility limit in a warmed solvent, and then letting it cool.
K s ) and are inherently unstable; dropping a "seed" crystal of the solid into such a solution will usually initiate rapid precipitation. But as is explained below, even a tiny dust particle may be enough. An old chemist's trick is to use the tip of a glass stirring rod to scrape the inner surface of a container holding a supersaturated solution; the minute particles of glass that are released presumably serve as precipitation nuclei.
The nucleation problem: precipitation is [theoretically] impossible!
Any process in which a new phase forms within an existing homogeneous phase is beset by the nucleation problem : the smallest of these new phases — raindrops forming in air, tiny bubbles forming in a liquid at its boiling point — are inherently less stable than larger ones, and therefore tend to disappear. The same is true of precipitate formation: if smaller crystals are more soluble, then how can the tiniest, first crystal, form at all?
In any ionic solution, small clumps of oppositely-charged ions are continually forming by ordinary collisional processes. The smallest of these aggregates possess a higher free energy than the isolated solvated ions, and they rapidly dissociate. Occasionally, however, one of these proto-crystallites reaches a critical size whose stability allows it to remain intact long enough to serve as a surface (a "nucleus") onto which the deposition of additional ions can lead to still greater stability. At this point, the process passes from the nucleation to the growth stage.
Theoretical calculations predict that nucleation from a perfectly homogeneous solution is a rather unlikely process; tenfold supersaturation should produce only one nucleus per cm 3 per year. Most nucleation is therefore believed to occur heterogeneously on the surface of some other particle, possibly a dust particle. The efficiency of this process is critically dependent on the nature and condition of the surface that gives rise to the nucleus.


8.8.3: Incarceration
Almost 2 million people are incarcerated in all of the state and federal prisons, jails, juvenile detention, and immigration detention centers in the United States (Sawyer & Wagner, 2023). At this point in time, Americans are incarcerated at a higher rate than anywhere else in the world. Since the 1990’s, rates of incarceration have increased exponentially, and only began to decline slightly in the last decade (see Figure 8.14 below). Yet, very little evidence supports the use of incarceration for improved public safety, and in fact, research suggests that incarceration has a substantial impact on the health of those impacted by the justice system, as well as their families and communities ( Incarceration - Healthy People 2030 , n.d.).
Many people are jailed before they have been convicted of a crime due to the bail system. The average amount of bail is $10,000, which many of those arrested cannot afford - and thus they await their day in court in pretrial detention. Additionally, almost 1 in 5 people incarcerated are there because of some violation of their parole - often a technicality (Sawyer & Wagner, 2023). Incarceration disproportionately affects people of color: 38% of the incarcerated population are Black Americans, and 21% are Hispanic - at much higher rates than the demographics of the general population (Wang et al., 2022).
Those living in prison or jail are more likely to be in poorer mental and physical health; including having higher rates of mental illness, substance use disorders, high blood pressure, asthma, cancer, and infectious diseases ( Incarceration - Healthy People 2030 , n.d.). And prisons are not great places to get treatment for mental health, addiction, or physical ailments either. Very few offer treatment programs for addiction, and suicide is one of the leading causes of death in prison. Even after release, formerly-incarcerated people often suffer from poor mental health and PTSD, on top of employment and other socioeconomic hurdles (Sawyer & Wagner, 2023).
The families and communities of those incarcerated are severely impacted as well. Nearly 113 million Americans have had an immediate family member spend at least 1 night in jail ( Every Second , n.d.). Children of incarcerated parents are more likely to have adverse childhood experiences - including having witnessed violence - and are more likely to live in poverty. Behavioral issues and learning challenges are also common, and these kids are at a higher risk of entering the system themselves ( Incarceration - Healthy People 2030 , n.d.). In fact, many of those incarcerated are victims of violence or abuse themselves - which demonstrates the vicious cycle described by the adage: “hurt people hurt people”. Higher fines and harsher sentences only seem to make this cycle worse and more difficult for impacted communities. High incarceration rates in a community only serve to disrupt social cohesion and decrease trust in government systems.


16.1: Introduction
In the past few chapters, specific types of infrastructure systems have been discussed. Infrastructure such as roadways and power generation is widespread and of great social importance. Organizations exist for the management of this infrastructure. While managers must consider inter-action and inter-dependency among different types of infrastructure, the institutional control over different infrastructure systems are typically independent.
Military bases, campuses, parks and ports are different with regard to institutional arrangements for infrastructure. Typically, a single organization is responsible for managing all of the infrastructure for these entities. Moreover, these integrated entities are generally sufficiently large to warrant employment of a professional staff of infrastructure managers. As a result, many professional infrastructure managers work in these integrated entities.
These facilities are also unusual in the types of infrastructure that may be included. Military bases have specialized infrastructure for their specific missions, such as ordnance storage facilities. Industrial and university campuses often have specialized laboratory equipment and facilities. Parks have historic buildings and priceless natural features. Ports have specialized equipment for handling freight and passengers. Even golf courses (considered here a form of parks) have special requirements for landscaping.
Figure 16.1.1 illustrates a former military base (Fort Jefferson) now located in a national park (Dry Tortugas) on an island remote from the mainland. This is a complicated facility, with historic structures for the fort and modern infrastructure to generate power, water supplies, wastewater and telecommunications. A small port is also on the island for both boats and seaplanes. As a national park, preservation of the natural eco-systems is a priority. Managing the infrastructure on this island is a complicated job for the National Park Service!
Figure 16.1.2 illustrates a university campus, which is another example of a single institution with multiple infrastructure systems. A university campus can rely on external infrastructures such as power generation and transportation services. But a multitude of services are provided on campus, including internal transportation circulation and parking, water and power distribution, and buildings.
Ports are another example of single-institution entities with a significant amount of infrastructure and specialized facilities. Airports, seaports, and inland waterways (as illustrated in Figure 16.1.3) are all examples of such entities.


1.2: Medical Language Related to the Body as a Whole
Learning Objectives
- Use the anatomic reference system to identify the anatomic position of the body
- Use the anatomic reference system to identify the body planes
- Use the anatomic reference system to identify the body cavities
- Use the anatomic reference system to identify the directional terms
- Use the anatomic reference system to identify the divisions of the body
- Describe the structural organization of the body
- Apply the rules of medical language
- Identify meanings of key word components
As you memorize the language components of medical terminology it is important to support that learning within the context of anatomy and physiology. Proceeding through the body system chapters you will learn word parts, whole medical terms, and common abbreviations. It is important to put into context where in the body the medical term is referencing, and then consider how it works within the body.
Anatomy focuses on structure and physiology focuses on function. Much of the study of physiology centers on the body’s tendency toward homeostasis .
Consider the structures of the body in terms of fundamental levels of organization that increase in complexity: subatomic particles, atoms, molecules, organelles, cells, tissues, organs, organ systems, organisms, and biosphere (Figure 2.1).
The Levels of Organization
All matter in the universe is composed of one or more unique pure substances called elements , familiar examples are hydrogen, oxygen, carbon, nitrogen, calcium, and iron.
-
The smallest unit of any of these pure substances (elements) is an
atom
.
- Atoms are made up of subatomic particles such as the proton, electron, and neutron.
-
Two or more atoms combine to form a
molecule
, such as the water molecules, proteins, and sugars found in living things.
- Molecules are the chemical building blocks of all body structures.
-
A
cell
is
the smallest independently functioning unit of a living organism.
- Even bacteria, which are extremely small, independently-living organisms, have a cellular structure. Each bacterium is a single cell. All living structures of human anatomy contain cells, and almost all functions of human physiology are performed in cells or are initiated by cells
- A human cell typically consists of flexible membranes that enclose cytoplasm, a water-based cellular fluid, together with a variety of tiny functioning units called organelles .
- A tissue is a group of many similar cells (though sometimes composed of a few related types) that work together to perform a specific function.
- An organ is an anatomically distinct structure of the body composed of two or more tissue types. Each organ performs one or more specific physiological functions.
An organ system is a group of organs that work together to perform major functions or meet the physiological needs of the body .
Did you know?
- Organs are very collaborative and work with multiple body systems.
- F or example, the heart (cardiovascular system) and lungs (respiratory system) work together to deliver oxygen throughout the body and remove carbon dioxide from the body.
Consider the breakdown into eleven distinct organ systems in the human body (Figure 2.2 and Figure 2.3). Assigning organs to organ systems can be imprecise since organs that “belong” to one system can also have functions integral to another system. In fact, most organs contribute to more than one system.
The organism level is the highest level of organization. An organism is a living being that has a cellular structure and that can independently perform all physiologic functions necessary for life. In multicellular organisms, including humans, all cells, tissues, organs, and organ systems of the body work together to maintain the life and health of the organism.
Watch this video:
Media 2.1. Introduction to Anatomy & Physiology: Crash Course A&P #1 [Online video]. Copyright 2015 by CrashCourse.
Anatomical Position
Health care providers use terminology for the purpose of precision and to reduce medical errors. For example, is a scar “above the wrist” located on the forearm two or three inches away from the hand? Or is it at the base of the hand? Is it on the palm-side or back-side? By using precise anatomical terminology, we eliminate ambiguity. Anatomical terms derive from ancient Greek and Latin words.
To further increase precision, anatomists standardize the way in which they view the body. Just as maps are normally oriented with north at the top, the standard body “map,” or anatomical position, is that of the body standing upright, with the feet at shoulder width and parallel, toes forward. The upper limbs are held out to each side, and the palms of the hands face forward as illustrated.
Using this standard position reduces confusion. It does not matter how the body being described is oriented, the terms are used as if it is in anatomical position. For example, a scar in the “anterior (front) carpal (wrist) region” would be present on the palm side of the wrist. The term “anterior” would be used even if the hand were palm down on a table.
A body that is lying down is described as either prone or supine. These terms are sometimes used in describing the position of the body during specific physical examinations or surgical procedures.
Regional Terms
The human body’s numerous regions have specific terms to help increase precision. Notice that the term “brachium” or “arm” is reserved for the “upper arm” and “antebrachium” or “forearm” is used rather than “lower arm.” Similarly, “femur” or “thigh” is correct, and “leg” or “crus” is reserved for the portion of the lower limb between the knee and the ankle. You will be able to describe the body’s regions using the terms from the anatomical position.
Directional Terms
Directional terms are essential for describing the relative locations of different body structures. For instance, an anatomist might describe one band of tissue as “inferior to” another or a physician might describe a tumor as “superficial to” a deeper body structure. Commit these terms to memory to avoid confusion when you are studying or describing the locations of particular body parts.
- Anterior (or ventral ) describes the front or direction toward the front of the body. The toes are anterior to the foot.
- Posterior (or dorsal ) describes the back or direction toward the back of the body. The popliteus is posterior to the patella.
- Superior (or cranial ) describes a position above or higher than another part of the body proper. The orbits are superior to the oris.
- Inferior (or caudal ) describes a position below or lower than another part of the body proper; near or toward the tail (in humans, the coccyx, or lowest part of the spinal column). The pelvis is inferior to the abdomen.
- Lateral describes the side or direction toward the side of the body. The thumb (pollex) is lateral to the digits.
- Medial describes the middle or direction toward the middle of the body. The hallux is the medial toe.
- Proximal describes a position in a limb that is nearer to the point of attachment or the trunk of the body. The brachium is proximal to the antebrachium.
- Distal describes a position in a limb that is farther from the point of attachment or the trunk of the body. The crus is distal to the femur.
- Superficial describes a position closer to the surface of the body. The skin is superficial to the bones.
- Deep describes a position farther from the surface of the body. The brain is deep to the skull.
Practice these directional terms.
Concept Check
- Find a partner and take turns choosing two body parts on your or your partner’s body.
- Using directional terms, describe the location of those body parts relative to one another.
Body Planes
A section is a two-dimensional surface of a three-dimensional structure that has been cut. Modern medical imaging devices enable clinicians to obtain “virtual sections” of living bodies. We call these scans. Body sections and scans can be correctly interpreted, however, only if the viewer understands the plane along which the section was made. A plane is an imaginary two-dimensional surface that passes through the body. There are three planes commonly referred to in anatomy and medicine:
- The sagittal plane is the plane that divides the body or an organ vertically into right and left sides. If this vertical plane runs directly down the middle of the body, it is called the midsagittal or median plane. If it divides the body into unequal right and left sides, it is called a parasagittal plane or less commonly a longitudinal section.
- The frontal plane is the plane that divides the body or an organ into an anterior (front) portion and a posterior (rear) portion. The frontal plane is often referred to as a coronal plane. (“Corona” is Latin for “crown.”)
- The transverse plane is the plane that divides the body or organ horizontally into upper and lower portions. Transverse planes produce images referred to as cross sections.
Can you locate the planes?
Body Cavities and Serous Membranes
The body maintains its internal organization by means of membranes, sheaths, and other structures that separate compartments. The dorsal (posterior) cavity and the ventral (anterior) cavity are the largest body compartments (Figure 2.6). These cavities contain and protect delicate internal organs, and the ventral cavity allows for significant changes in the size and shape of the organs as they perform their functions. The lungs, heart, stomach, and intestines, for example, can expand and contract without distorting other tissues or disrupting the activity of nearby organs.
Subdivisions of the Posterior (Dorsal) and Anterior (Ventral) Cavities
The posterior (dorsal) and anterior (ventral) cavities are each subdivided into smaller cavities:
The posterior (dorsal) cavity has two main subdivisions:
-
In the posterior (dorsal) cavity, the
cranial cavity
houses the brain
- Protected by the bones of the skulls and cerebrospinal fluid
-
The
spinal cavity
(or vertebral cavity) encloses the spinal cord.
- Protected by the vertebral column and cerebrospinal fluid
The anterior (ventral) cavity has two main subdivisions:
-
The
thoracic cavity
is the more superior subdivision of the anterior cavity, and it is enclosed by the rib cage.
- The thoracic cavity contains the lungs and the heart, which is located in the mediastinum.
- The diaphragm forms the floor of the thoracic cavity and separates it from the more inferior abdominopelvic cavity.
-
The
abdominopelvic cavity
is the largest cavity in the body.
- No membrane physically divides the abdominopelvic cavity.
- The abdominal cavity houses the digestive organs, the pelvic cavity, and the reproductive organs.
Practice locating cavities.
Abdominal Regions and Quadrants
To promote clear communication, for instance about the location of a patient’s abdominal pain or a suspicious mass, health care providers typically divide up the cavity into either nine regions or four quadrants.
Practice locating the quadrants.
Tissue Membranes
A tissue membrane is a thin layer or sheet of cells that covers the outside of the body (for example, skin), the organs (for example, pericardium), internal passageways that lead to the exterior of the body (for example, abdominal mesenteries), and the lining of the movable joint cavities. There are two basic types of tissue membranes: connective tissue and epithelial membranes (Figure 2.7).
Connective Tissue Membranes
-
The
connective tissue membrane
is formed solely from connective tissue.
- These membranes encapsulate organs, such as the kidneys, and line our movable joints.
-
A
synovial membrane
is a type of connective tissue membrane that lines the cavity of a freely movable joint.
- For example, synovial membranes surround the joints of the shoulder, elbow, and knee.
Epithelial Membranes
-
The
epithelial membrane
is composed of epithelium attached to a layer of connective tissue.
- For example, your skin.
-
The
mucous membrane
is also a composite of connective and epithelial tissues.
- Sometimes called mucosa, these epithelial membranes line the body cavities and hollow passageways that open to the external environment, and include the digestive, respiratory, excretory, and reproductive tracts.
- Mucus, produced by the epithelial exocrine glands, covers the epithelial layer.
- The underlying connective tissue, called the lamina propria (literally “own layer”), help support the fragile epithelial layer.
-
The skin is an epithelial membrane also called the
cutaneous membrane
.
- It is a stratified squamous epithelial membrane resting on top of connective tissue. The apical surface of this membrane is exposed to the external environment and is covered with dead, keratinized cells that help protect the body from desiccation and pathogens.
Membranes of the Anterior (Ventral) Body Cavity
-
A
serous membrane
(also referred to as serosa) is an epithelial membrane composed of mesodermally derived epithelium called the mesothelium that is supported by connective tissue.
- Parietal layers: line the walls of the body cavity.
- Visceral layer: covers the organs (the viscera).
- Between the parietal and visceral layers is a very thin, fluid-filled serous space.
There are three serous cavities and their associated membranes. Serous membranes provide additional protection to the viscera they enclose by reducing friction that could lead to inflammation of the organs.
- Pleura : surrounds the lungs in the pleural cavity and reduces friction between the lungs and the body wall.
- Pericardium: surrounds the heart in the pericardial cavity and reduces friction between the heart and the wall of the pericardium.
- Peritoneum: surrounds several organs in the abdominopelvic cavity. The peritoneal cavity reduces friction between the abdominal and pelvic organs and the body wall.
Test Yourself
References
[CrashCourse]. (2015, January 6). Introduction to anatomy & physiology: Crash course A&P #1 [Video]. YouTube. https://youtu.be/uBGl2BujkPQ
Unless otherwise indicated, this chapter contains material adapted from Anatomy and Physiology (on OpenStax ), by Betts, et al. and is used under a a CC BY 4.0 international license . Download and access this book for free at https://openstax.org/books/anatomy-and-physiology/pages/1-introduction .


2.11: Assessment of Cardiovascular Function
Introduction to the cardiac cycle
Several physical and electrical events occur during the cardiac cycle , resulting in the flow of blood through the heart and peripheral tissues. In the simplest terms, the cardiac cycle traces the flow of blood through the heart. The physical contraction and relaxation of atria and ventricles is induced by the proper functioning of the heart’s electrical conduction system .
At the beginning of the cardiac cycle, the atria are relaxed and fill with blood coming from the vena cavae and cardiac venus sinus or the pulmonary veins. The increased pressure in the atria due to atrial filling eventually forces the atrioventricular (AV) valves open, and blood flows into the ventricles. The atria contract and force the remaining blood into the ventricles, which are relaxed during filling.
As the pressure of the blood inside the ventricles increases, the atrioventricular valves close. During this time the arterial pressure is at its lowest point during the cardiac cycle; this is called diastolic pressure. Ventricular contraction increases the ventricular pressure until it exceeds the arterial pressure, forcing the semilunar valves open, causing blood to flow into the pulmonary arteries and the aorta. Blood entering the arterial system from the aorta inflates the arteries and increases blood pressure to a maximum, which is the systolic pressure .
In this lab you will record and electrocardiogram (ECG) from a subject and listen to the characteristic “lub-dub” heart sounds. The “lub” sound occurs during the early phase of ventricular contraction and is produced by closing of the atrioventricular valves, which prevents blood backflow into the atria. When the ventricles relax, the blood pressure drops below what is in the arteries and the semilunar valves close, producing the “dub” sound. These sounds can be correlated with the basic waveforms seen on an ECG (Figure 1).
Electrical conduction system of the heart
The heart is composed of two basic types of cells – contractile and non-contractile. Most cardiomyocytes are contractile, and when they are stimulated by an action potential, they contract. Some cardiomyocytes are non-contractile: that is, they only conduct action potentials to the contractile cells. These specialized cardiomyocytes are called conducting fibers (Figure 2).
Some of the conducting cells in the heart normally generate action potentials on their own (without nervous system stimulation). These cells, which lead to the autorhythmic nature of the heart , control the timing of contraction in the atria and ventricles. The most important autorhythmic cells of the heart are found in the sinoatrial (SA) node and the atrioventricular (AV) node . The pacemaker of the heart, which sets the overall heart rhythm in normal individuals, is the SA node. The SA node generates and transmits signals, stimulating the atria to contract and sending an impulse to the AV node.
The signal sent from the SA node to the AV node informs the AV node cells that the atria have contracted. The AV node delays transmission of a secondary impulse to the rest of the heart. This short delay, normally 0.12 seconds, is very important because it allows enough time for complete atrial contraction before the ventricles contract (that is, enough time for the ventricles to finish filling with atrial blood). This delay coordinates the successive contractions of the atria and ventricles.
From here, the AV node sends an impulse through the Bundle of His , which quickly branches into the right and left interventricular bundle branches . These branches send the impulse through the ventricular septum of the heart. Once the impulse reaches the apex of the heart , ventricular contraction ensues. The two bundle branches taper to produce the Purkinje fibers , which stimulate the myocardial cells composing the ventricular walls to contract. Purkinje fibers first conduct the impulse to contractile cardiomyocytes at the apex and then to the cardiomyocytes of the lateral walls of the heart. The result is that the ventricles contract from the bottom upward, starting contraction at the apex and ending contraction in the regions of the ventricles proximal to the atria. This method of contraction allows for adequate squeezing of the blood out of the ventricles and into their respective arteries (either the pulmonary artery or aorta), much like squeezing a tube of toothpaste from the bottom up in order to force out all the toothpaste.
Characteristics of the ECG
The electrical activity (action potentials) generated by the contractile cardiomyocytes of the heart during atrial and ventricular contraction can be measured by an electrocardiogram (ECG) . The most-simple ECG can be recorded by placing electrodes on the two wrists and left ankle, forming Einthoven’s Triangle . ECG electrodes record the electrical potentials originating from the cells of the heart as impulses spread from the atria to the ventricles. These impulses cause very small electrical changes in the skin that occur in response to cardiac action potentials.
Although the timing and amplitudes for normal findings in the ECG may vary, the overall shape and direction (deflection) of ECG waveforms will be similar under normal physiological conditions (Figure 1, above). Combining ECG information with a patient’s overall clinical status makes it an extremely valuable method for determining heart function. Analysis of ECGs includes examination of ECG waves, intervals (a section of the ECG that contains a wave), and segments (an isoelectric, or flat, section of the ECG that is between two sequential waves).
The P-wave represents depolarization of the plasma membranes of cardiomyocytes leading to the sequential contraction of the right and left atria. Sometimes it is possible to see this sequential activation as a biphasic wave representing the spread of charge from the right to the left side of the heart. The duration of the P-wave is typically no greater than 0.12 seconds.
The QRS complex represents depolarization leading to the simultaneous contraction of cardiomyocytes of the right and left ventricles. Most of the electrical activity recorded by the QRS wave is derived from the left ventricle, which is larger and more muscular. The duration of the QRS complex is typically 0.06-0.10 seconds. The amplitude of the QRS complex may vary, depending on the size of the ventricles and the proximity of the electrodes to the ventricular chambers.
The T-wave represents the repolarization of the ventricles, which results in muscle cell relaxation.
Note: the repolarization wave of atrial cell membranes resulting in atrial relaxation is hidden within the QRS complex.
Measuring the time elapsed between ECG features can provide useful information when diagnosing heart abnormalities (Figure 3, and Table 1 below):
The PR interval is the region of the graph that extends from the beginning of the P wave (onset of atrial depolarization) until the start of the QRS complex (onset of ventricular depolarization). The duration of this interval is normally between .12-.20 seconds. The relatively slow conduction through this interval is important because it ensures that atrial and ventricular depolarizations are separated in time; thus, allowing adequate time for ventricular filling before the ventricles contract.
Pathologies of the Conduction System of the Heart: Heart Block
Heart block refers to the conduction of impulses between the atria and ventricles that is too slow or non-existent. Long PR intervals can indicate first degree heart block , which is caused by conduction through the AV node that is too slow. Slow conduction through the AV node is called AV delay . AV delay may be caused by damage to the AV node tissues due to myocardial infarction (heart attack).
When AV conduction is completely blocked, the ECG displays back-to-back P waves. When P waves are not followed by the QRS complex, the ventricles fail to contract, and this leads to reduced mean arterial pressure (MAP) that can cause dizziness and fainting (due to the drop in cerebral blood pressure). When the ECG trace displays an occasional extra P-wave, where, otherwise, impulses still travel through the AV node , second degree heart block is diagnosed.
In the most severe case, P waves and QRS complexes may become independently regulated by the heart’s conducting tissues because impulses never pass through the AV node. This condition leads to the atria and ventricles having their own distinct sinus rhythm, called third degree heart block . In this case, the ECG may show several extra P waves before a QRS complex.
The Q-T interval , as its name suggests, measures the time from the start of the Q wave to the end of the T wave. This measures the time for depolarization and repolarization, or contraction and relaxation, of the heart’s ventricles. The Q-T interval is dependent upon heart rate: a faster heart rate results in a shorter Q-T interval. Reported values for normal QT intervals in resting individuals are equal to or lesser than 0.43 seconds. Longer than average Q-T intervals might indicate ventricular arrhythmias, depending on the resting heart rate and patient activity.
The R-R interval is similar to the Q-T interval in that it is also dependent upon heart rate. The R-R interval is the latency between QRS complexes, or the time between successive depolarizations of the ventricles. The R-R interval measures the periodicity of the heart beat: that is, it measures the regularity of the heart beat. The R-R interval can be used to calculate heart rate .
The normal R-R interval is between 0.6 seconds to 1 second, and correlates to an average resting heart rate of 60-100 bpm. Heart rates vary in adults depending upon age, weight, fitness and other factors. Normal heart rates range from 60-100 bpm, but in highly trained athletes, the rate may be lower (40-60 bpm). Heart rates lower than 60 bpm are considered bradycardia, or abnormally slow. Heart rates over 100 bpm are considered tachycardia , or abnormally fast.
For a normal heart rate, we consider the conduction velocity of the myocardial cells of the conduction system of the heart. Just as we measured the conduction velocity of the ulnar nerve in the Week 7 laboratory, myocardial conducting tissues have their own normal conduction velocities, even though they are not neural tissue. That is, the speed at which action potentials are propagated within the tissue from one site to the next can be calculated.
The conduction velocity in the AV node is the slowest of all the myocardial tissues (0.01-0.05 m/s). compare this value from the AV node with the conduction velocities for other myocardial tissues: atrial and ventricular conduction velocities average 1 m/s and the Bundle of His and Purkinje fibers average 2-4 m/s.
| ECG Feature | Duration (seconds) | Description |
|---|---|---|
| P Wave | .08-0.10 s | atrial depolarization |
| QRS Complex | .08-0.10 s | ventricular depolarization |
| T Wave | 160 ms | ventricular repolarization |
| PR Interval | 0.12-0.2 s | time between onset of atrial depolarization and ventricular depolarization |
| RR Interval | 0.6-1.2 s | rate of ventricular cycle |
| PP Interval | 0.6-1.2 s | rate of atrial cycle |
| QT Interval | .40-.43 s | time required for ventricular depolarization and repolarization |
| Heart Rate | 60-100 BPM |
Laboratory Methods
Conducting the Electrocardiogram and its Analysis
Equipment Required
iWire-B3G ECG cable and electrode lead wires, EM-220 Event marker, Stethoscope, Alcohol swabs, Disposable ECG electrodes
Start the Software
- Turn on the iWorx unit with the switch at the back of the box.
- Locate the P-drive and double click the Week 10 settings file to open LabScribe.
Set Up
- Instruct the subject to remove all jewelry from their wrists and ankles . Use an alcohol swab to clean a region of skin on the inside of the subject’s right wrist, left wrist and left ankle. Let the area dry.
- Remove a disposable ECG electrode pad from its plastic shield and apply the electrode pad to the scrubbed area
- Snap the lead wires onto the electrodes, so that:
- The RED (+1) lead is attached to the left wrist
- THE BLACK (-1) lead is connected to the right wrist,
- THE GREEN (C or ground) lead is connected to the left ankle.
- Instruct the subject to sit quietly with their hands in their lap holding the iWire B3G box . If the subject moves, the ECG trace will move off the top or bottom of the screen. If the subject moves any muscles in the arms or upper body, electromyograms (EMGs) from the muscles will appear on the ECG recording as noise.
If in Exercise 1 you do not see a normal ECG recording, you may relocate the electrodes from Lead I to Lead II or Lead II (see figures 1a and 1b below).
| Lead | Red (positive) | Black (negative) | Green (ground) |
|---|---|---|---|
| I | Left arm , distal to the antecubital area | Right arm , distal to the antecubital area | Inside left ankle , behind the medial malleolus |
| II | Inside of left ankle , behind the medial malleolus | Right arm , distal to the antecubital area | Inside right ankle , behind the medial malleolus |
| III | Inside left ankle , behind the medial malleolus | Left ar m, distal to the antecubital area | Inside right ankle , behind the media malleolus |
Exercise 1: The ECG in a Resting Subject
Aim: To measure the ECG in a resting individual.
Procedure
- Click on the Record button , located on the upper right side of the LabScribe3 Main window. The signal should begin scrolling across the screen.
- Click on the AutoScale button at the upper margin of the ECG channel. If the signal on the ECG channel is upside down compared to methods figure 2 (below), click on the downward arrow to the left of the channel title and select the Invert function. The trace should now look similar to the one in the figure.
3. When you have a suitable trace, type <Subject’s Name> Resting ECG in the Mark box to the right of the Mark button. Press the Enter key on the keyboard to attach the comment to the data. Record for a minute or two. Watch the time in the upper right-hand corner of the screen.
4. Click Stop to halt recording.
5. Select Save As in the File menu, type a name for the file. Choose a destination on the computer in which to save the file, like your lab group folder. Click on the Save button to save the data file.
Data Analysis
- Scroll through the recording and find a section of data with 4 to 6 optimal ECG cycles in succession.
- Use the mouse to click on and drag the cursors to specific points on the ECG recording to measure the following:
-
- The R-wave amplitude. To measure the R wave amplitude , place one cursor on the Q wave that precedes the R wave and the second cursor on the peak of the R wave . The value for V2-V1 on the ECG channel is this amplitude. Measure the amplitudes of two additional R waves in mV.
- The P-wave amplitude. To measure the P wave amplitude , place one cursor on the baseline that precedes the P wave and the second cursor on the peak of the P wave . The value for V2-V1 on the ECG channel is this amplitude. Measure the amplitudes of two additional P waves in mV.
- The T-wave amplitude. To measure the T wave amplitude , place one cursor on the baseline that precedes a P wave and the second cursor on the peak of the T wave that is in the same cycle as that P wave . The value for V2-V1 on the ECG channel is this amplitude. Measure the amplitudes of two additional T waves in mV .
- The beat period , which is the time interval between two adjacent R waves (R-R interval) . To measure the beat period , place one cursor on the peak of a R wave and the second cursor on the peak of the adjacent R wave. The value for T2-T1 on the ECG channel is the beat period. Measure the beat period for two additional pairs of R waves in seconds.
- Record the amplitude and time interval values in your lab report.
- Calculate the following values and record on your lab report:
- The average amplitudes for ECG waves and average beat period, in beats per second .
- The heart rate, which is expressed in beats per minute and calculated from the average beat period by using the following equation:
Heart Rate (beats/minute) = (60 seconds/minute) / (# seconds/beat)
Exericse 2: The ECG and Heart Sounds
Aim: To study the relationship of heart sounds to the ECG.
Procedure
- Place the head of the stethoscope on the left side of the subject’s chest and listen for the heart sounds . Move the stethoscope head to the different auscultation positions shown above. Decide which position has the best-defined heart sounds.
- Click on the Record button ( you will record for 30 seconds) . With one hand, hold the stethoscope head on the subject’s chest where the “lub-dub” is heard the best, and hold the event marker in the other hand. Press the event marker when you hear the “ lub ”, or first heart sound, and release it when you hear the “dub ”, or second heart sound. This will happen very quickly! You should practice several times.
- After recording for at least twenty seconds , click Stop to halt recording. After several tries, if the event marker for the “Lub” does not occur after the R-Wave and the event marker for the “Dub” does not align with or come right after the T-Wave, ask your TA for help.
- Select Save in the File menu on the LabScribe3 window.
Data Analysis
- Scroll through the recording and find a section of data with 4 to 6 exemplary ECG waveforms and consistent responses on the event marker channel, in succession.
Use the mouse to click on and drag the cursors to specific points on the ECG recording to measure the following:
Note: Do the measurement for the R-Lub Interval and the T-Dub Interval on the same wave, making sure the event marker for the Lub is after the R-wave and the event marker for the Dub is after the T-Wave (Methods Figure 4).
- The R-Lub Interval , which is the time interval between the peak of a R wave and the onset of the event marker . The onset of the event marker indicates the occurrence of the first heart sound or “lub”. Record the value for T2-T1 in seconds. Measure this time interval for two additional ECG cycles .
- The T-Dub Interval , which is the time interval between the peak of a T wave and the offset of the event marker. The offset of the event marker indicates the occurrence of the second heart sound or “dub”. Record the value for T2-T1 in seconds . Measure this time interval for two additional ECG cycles .
- Calculate the following values and record on your data table: the average R-Lub interval in seconds; the average T-Dub interval in seconds .